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805+ (Hard)|   Combinations|   Geometry|      
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Bunuel
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How many rectangles can be formed by taking the four vertices of 18-si 04 Feb 2020, 03:41
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How many rectangles can be formed by taking the four vertices of 18-sided regular polygon?

A. 16
B. 24
C. 32
D. 36
E. 3060

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exc4libur
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How many rectangles can be formed by taking the four vertices of 18-si Updated on: 05 Feb 2020, 05:38
Originally posted by exc4libur on 04 Feb 2020, 08:40.
Last edited by exc4libur on 05 Feb 2020, 05:38, edited 2 times in total.
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Bunuel
How many rectangles can be formed by taking the four vertices of 18-sided regular polygon?

A. 16
B. 24
C. 32
D. 36
E. 3060
Show more

straight lines: C(18,2)
triangles: C(18,3)
quadrilaterals: C(18,4)=18!/4!14!=3060
rectangle (4 right angles): 9!/2!=36

imagine the polygon in half and choose any two-points, from that half;
now, the number of ways we can choose the other two-points from opposite half is 1;
so, 18-sided polygon's half has 9 vertices to choose any two-points from: 9C2
and the number of ways to select the opposite vertices: 1
total: 9C2*1=36

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rajatchopra1994
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How many rectangles can be formed by taking the four vertices of 18-si Updated on: 05 Feb 2020, 03:47
Originally posted by rajatchopra1994 on 04 Feb 2020, 03:47.
Last edited by rajatchopra1994 on 05 Feb 2020, 03:47, edited 1 time in total.
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Thanks Bunuel ,

Explanation:

Let take an example for Hexagon, it will 3 rectangles (with 3 Diagonal pairs which are parallel)

9C2 = 36

IMO-D
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Bunuel
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How many rectangles can be formed by taking the four vertices of 18-si 05 Feb 2020, 03:15
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This is not an easy question. Notice that we need rectangles, not simply quadrilaterals.
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Subhrajyoti
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How many rectangles can be formed by taking the four vertices of 18-si 14 Feb 2020, 07:22
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18 sided polygon will have 18 vertices, using combination i.e. C( 18,4) = 3060 E
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How many rectangles can be formed by taking the four vertices of 18-si 26 May 2020, 16:43
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exc4libur
Bunuel
How many rectangles can be formed by taking the four vertices of 18-sided regular polygon?

A. 16
B. 24
C. 32
D. 36
E. 3060

straight lines: C(18,2)
triangles: C(18,3)
quadrilaterals: C(18,4)=18!/4!14!=3060
rectangle (4 right angles): 9!/2!=36

imagine the polygon in half and choose any two-points, from that half;
now, the number of ways we can choose the other two-points from opposite half is 1;
so, 18-sided polygon's half has 9 vertices to choose any two-points from: 9C2
and the number of ways to select the opposite vertices: 1
total: 9C2*1=36

Ans (D)
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Hi, can't this also be done by something like this:
(18*17)/4

18 because can select any of the 18 vertices and 17 because after selecting the vertex, one can choose any of the remaining 17 vertices and then multiply by 1 because for the remaining 2 vertices there is just one possibility.

Divide by 4 for repetition, please let me know where I'm going wrong
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How many rectangles can be formed by taking the four vertices of 18-si 06 Jun 2020, 09:30
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For making a rectangle we need 4 points ( at 90 deg angle when connected)
so we have to divide the polygon with 9 points on one side ( say 1,2,3..9) and 9 on the other
Now, from this 9 points
first select point 1, the combinations can be (1,2) (1,3) (1,4) ....(1,9) -----> 8 sets of rectangle
Note that we have the corresponding other points in the other 9 points (10, 11, ... 18)
Total we have 8 rectangle

similarly, for point 2, the combinations can be (2,3) (2,4) ...(2,9) ------> 7 sets of rectangle
same way we will have for all the points up to point 8, for which we have 1 rectangle

8+7+...1 =36
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How many rectangles can be formed by taking the four vertices of 18-si 19 Jul 2024, 01:20
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